All categories

3 years ago

Please help!!!

Physics

1 answer:

Brilliant_brown [7]3 years ago

5 0

Answer:

All of the inventors can be characterized as problem-solvers.

Explanation:

Took the practice

You might be interested in

A large water tank is 3.70 m high and filled to the brim, the top of the tank open to the air. A small pipe with a faucet is att

ipn [44]

h =(3.7 - .58)m = 3.12m

Now put PE into KE and we have to use the formula:

√2gh (g = gravity and h = height) therefor:

√2 x 9.8 x 3.12

= 7.82m/s

I hope this helps!

7 0

3 years ago

S Suppose you wish to fabricate a uniform wire from a mass m of a metal with density rhom and resistivity rho. If the wire is to

denpristay [2]

The diameter of the wire is 2.8 * 10^-3 m.

<h3>What is the length?</h3>

Mass of the wire = 1.0 g or 1 * 10^-3 Kg

Resistance = 0.5 ohm

Resistivity of copper = 1.7 * 10^-8 ohm meter

Density of copper = 8.92 * 10^3 Kg/m^3

V = m/d

But v = Al

Al = m/d

A = m/ld

Resistance = ρl/A

= ρl/m/ld =

l^2 = Rm/ρd

l = √ Rm/ρd

l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3

l = 1.82 m

A = πr^2

Also;

A = m/ld

A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3

A = 6.2 * 10^-5 m^2

r^2 = A/ π

r = √A/ π

r = √6.2 * 10^-5 m^2/3.142

r = 1.4 * 10^-3 m

Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m

Learn more about resistivity:brainly.com/question/14547003

#SPJ4

Missing parts;

Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?

5 0

2 years ago

Why is radiation often used to destroy cancer cells ?

zimovet [89]

Answer:

Explanation:

Radiation affects both cancer cells and healthy cells, but it affects cancer cells more.

8 0

3 years ago

How must the height of A relate to the height of B for the skateboarder to have just enough energy to complete the loop?

KiRa [710]

Answer:

The answer is D

Explanation:

8 0

3 years ago

Read 2 more answers

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago